LeetCode（Array)1480. Running Sum of 1d Array

骑猪看日落 2023-09-23 22:58 62阅读 0赞

# 1.问题 #

Given an array nums. We define a running sum of an array as runningSum\[i\] = sum(nums\[0\]…nums\[i\]).

Return the running sum of nums.

Example 1:

> Input: nums = \[1,2,3,4\]  
> Output: \[1,3,6,10\]  
> Explanation: Running sum is obtained as follows: \[1, 1+2, 1+2+3, 1+2+3+4\].

Example 2:

> Input: nums = \[1,1,1,1,1\]  
> Output: \[1,2,3,4,5\]  
> Explanation: Running sum is obtained as follows: \[1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1\].

Example 3:

> Input: nums = \[3,1,2,10,1\]  
> Output: \[3,4,6,16,17\]

Constraints:

*  1 <= nums.length <= 1000
 *  \-10^6 <= nums\[i\] <= 10^6

# 2. 解题思路 #

## 方法1： ##

> 1.定义result数组长度和nums相同，定义int sum初始为0  
> 2.遍历nums数组，sum+=nums\[i\]是遍历元素后相加的值，值添加到result  
> 3.返回result

## 方法2： ##

> 在原来的nums数组修改元素，nums的初始为nums\[0\]不用计算，输出结果的nums\[1\]=nums\[0\]+nums\[1\],以此遍历循环，求出修改后的nums数组

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public int[] runningSum(int[] nums) {
        
            int[] result = new int[nums.length];//1.定义result数组长度和nums相同，定义int sum初始为0
            int sum =0;
            for (int i = 0; i < nums.length; i++) {
        //2.遍历nums数组，sum+=nums[i]是遍历元素后相加的值，值添加到result
                result[i]=(sum+=nums[i]);
            }
            return result;   //3.返回result
        }
    }

## 代码2： ##

public int[] runningSum(int[] nums) {
        
            for(int i = 1 ; i < nums.length ; i++){
        //在原来的nums数组修改元素，nums的初始为nums[0]不用计算，输出结果的nums[1]=nums[0]+nums[1],以此遍历循环求出修改后的nums数组
                nums[i]=nums[i-1]+nums[i];
            }
            return nums;
        }